Solving a Student's Experiment on an Ideal Gas: Molar Specific Heat Capacity

In summary: For part (ii), $$C_V = \frac{Q}{n \Delta T} = \frac{Q R}{P \Delta V} = \frac{Q R}{n \Delta T} = C_p$$since ##\Delta T = \Delta U##, the change in internal energy.In summary, for an ideal gas, the molar specific heat capacities at constant pressure and constant volume are equal. This can be derived from the ideal gas law and the first law of thermodynamics.
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Homework Statement



A student performs an experiment on an ideal gas by adding 42.0 J of heat to it. As a result the student finds that the volume of the gas changes from 50 cm3 to 150 cm3 while the pressure remains constant at 101.3 kPa.

i) If the quantity of the gas present is 0.007 moles, determine the molar specfic heat capacity of the gas that the student would find at constant pressure.

ii) What is the molar specific heat capacity of the gas at constant volume?

2 Relevant equations

For Gases:

PV = nRT

Q = n . Cv . ΔT for constant-volume processes
Q = n . Cp . ΔT for constant-pressure processes

Cp = Cv + R

ΔU = Q - W

W = P . ΔV for an isobaric process

Q = 42J
Vi = 50 cm3 = 50 μm3
Vf = 150 cm3 = 150μm3
P = 101300 Pa
n = .007 moles
R = 8.315 J / (mol K)


The Attempt at a Solution



First, I worked out the work done on the system by the surroundings:

Since this is an isobaric expansion,

W = P . ΔV
= 101300 * (150 - 50)μ
= 10.13 J

so ΔU is;

ΔU = Q - W
= 42 - 10.13
= 31.87 J.

I am not sure why I calculated the change in internal energy. Do I even use it in this whole process?


"Q (constant pressure) must account for both the increase in internal energy and the transfer of energy out of the system by work"

for (i) the molar specific heat capacity at a constant pressure,

Q = n . Cp . ΔT

so Cp = Q / ( n . ΔT )


and PV = nRT

so T = ( PV ) / ( nR)

=> ΔT = Tf - Ti
= { ( PVf ) / ( nR) } - { ( PVi ) / ( nR) }
= ( P / nR ) ( Vf - Vi)
= ( 101300 / (0.007 * 8.315) ) ( (150-50)μ)
= 174.0400 K

So Cp is;

Cp = Q / ( n . ΔT )
= 42 / ( 0.007 * 174.0400)
= 34.47482725 J mol-1 K -1

Therefore the molar specific heat capacity of the gas at a constant pressure of 101300 Pa is 34.5 J mol-1 K -1

I think that that might be the right answer, but is my reasoning correct? I'm not too sure since I have been spending an age trying to work this one out and I keep getting different values, I have approached several "classmates" who turn their backs at the call for help, but typing slowy and going through this whole process of listing everyting down, I think I may have finally cracked it. Can you please verify this.
*For all the PHY160s out there who can't find a willing helper, this one is for you. I sincerely do hope that all of you, even those who would prefer not to share methods, get that 2% from this online assesment which would be so ever useful in your bid to get invited to the Medical Interviews*



As for (ii) What is the molar specific heat capacity of the gas at constant volume?


I am not sure how I can solve this using:

Q = n . Cv . ΔT for constant-volume processes

However, after patiently churning through my notes, admist all the equations and explanations in monochrome, I found the formula and note:

Cp = Cv + R "NB: This is true for ALL conditions."

So Cv is;

Cv = Cp - R
= 34.47482725 J mol-1 K -1 - 8.315 J mol-1 K -1
= 26.15982725
= 26.2 J mol-1 K -1

Thefore the molar specific heat capacity of the gas at constant volume is 26.2 J.

This answer also agrees with the provided answer. I know that the note says that "this is true for all conditions" so does this mean that it is true for, even, this condition?

And is there any other way to solve a problem like this?


Thanks for all your help. Even writing this post has helped me alot!
 
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  • #2
The OP's solutions for both parts appear to me to be correct.

There is no need to calculate the temperature change. For a constant pressure process of an ideal gas, the ideal gas law yields ##n \Delta T =\frac { P \Delta V}{R}##. So, $$C_p = \frac{Q}{n \Delta T} = \frac{Q R}{P \Delta V}$$
 

FAQ: Solving a Student's Experiment on an Ideal Gas: Molar Specific Heat Capacity

What is an ideal gas?

An ideal gas is a theoretical gas made up of particles that have negligible volume and exert no intermolecular forces on each other. This means that the particles are considered to be point masses that move independently and collide elastically with each other and the walls of the container.

What is molar specific heat capacity?

Molar specific heat capacity is the amount of heat required to raise the temperature of one mole of a substance by one degree Celsius. It is expressed in units of J/mol·K (Joules per mole Kelvin).

How do you solve a student's experiment on an ideal gas?

To solve a student's experiment on an ideal gas, you first need to collect data on the pressure, volume, and temperature of the gas. Then, you can use the ideal gas law (PV=nRT) to calculate the number of moles of gas present. Finally, you can use the molar specific heat capacity equation (q=nCΔT) to calculate the heat absorbed or released by the gas.

What factors can affect the accuracy of the results in a student's experiment?

There are several factors that can affect the accuracy of the results in a student's experiment, including experimental errors, variations in the ideal gas conditions, and the use of an imperfect measuring instrument. Additionally, the assumption that the gas behaves ideally may not hold true in certain situations.

How can the results of a student's experiment on an ideal gas be applied in real-world situations?

The results of a student's experiment on an ideal gas can be applied in real-world situations to determine the molar specific heat capacity of a gas under various conditions. This information can be useful in industries such as engineering, chemistry, and physics for designing and optimizing processes involving gases. It can also be used to better understand and predict the behavior of gases in different environments.

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